Mechanisms
A reaction mechanism is a set of simple, single, elementary steps that together sum to a chemical reaction. Rate-Determining Steps Naturally, some elementary steps will be slower, or even much slower than others due to a higher required activation energy. In a similar manner as limiting reactants, some of these steps will limit how fast later steps can go. Say you have A + B --> C slow C + A --> E fast E cannot be produced until C is produced by the first step. Since the first step is much slower, the fast step cannot proceed until the slow step has occurred. Essentially, once C is produced, finishing the reaction is almost instantaneous. Therefore, the first step is a rate-determining step. 'They are important because they determine the rate of the entire reaction. Furthermore, the rate law for the rate-determining step is the same as the rate law for the entire reaction because it limits the rate of the reaction. This leads to the question of how to write this rate law. Because steps are elementary and only involve one, two, or three particles reacting, collision theory can be applied without having to rely on empirical data. Criteria for Proposed Mechanisms #The steps must add up to the overall balanced equation #Elementary steps must be reasonable (unimolecular or bimolecular) #The proposed mechanism must agree with the experimentally determined rate law Note that mechanisms can never be proven with these steps because there could be another equally valid one. Mechanisms can only be refuted by violating one of these criteria. Determining Rate Laws from Mechanisms Knowing which step is the rate-determining step is a crucial step in order to determine the overall rate law. Refer to the previous section for review on how to do this. Mechanisms with the rate-determining step first are much easier, so I'll start with those. Slow initial Step Say you have the reaction: NO2(g) + CO(g) -> NO(g) + CO2(g) With a proposed mechanism: NO2 + NO2 -> NO + NO3 NO3 + CO -> NO2 + CO2 Despite previous premonitions that rate laws can only be determined with empirical data, mechanisms allow us to determine a rate law without an experiment. First determine that the first step is slow because nitrogen trioxide from the first step is required as a reactant in the second step. Next, write the rate laws of both steps. rate 1 = k1NO22 rate 2 = k2NO3CO Since the first step is the rate-determining step, its rate law is the overall rate law. Fast initial Step Finding the rate law of a reaction mechanism with a fast initial step involves similar steps as previously described with the slow initial step. Since the fast initial steps build up product and cannot be removed until the slow step removes it, some product is converted back to reactants because of le chatelier's principles and establishes an equilibrium. Therefore, if a step is reversible, it must be fast. Consider this example, hydrogen gas reacting with nitric oxide. 2H2(g)'' + 2NO''(g)'' --> 2H2O''(g)'' + N2''(g)'' The proposed mechanism: (1) 2NO''(g)'' <--> N2O2''(g)'' (2) 2H2''(g) <--> 4H(g)'' (3) N2O2''(g)'' + H''(g)'' à N2O''(g)'' + HO''(g)'' (4) 2HO''(g)'' + 2H''(g)'' à 2H2O''(g)'' (5) H''(g)'' + N2O''(g)'' à HO''(g)'' + N2''(g)'' So, first determine the rate-determining step. Because the first step is reversible, we know that dinitrogen dioxide will build up due to a slow step with a reactant of dinitrogen dioxide. The only step with dinitrogen dioxide as a reactant is the third step, so we know this is the slow step, or rate-determining step. So, let's write rate laws for each step. Rate 1 = k1NO2 Rate 2 = k2H22 '''Rate 3 = k3N2O2H Rate 4 = k4HO2H2 Rate 5 = k5HN2O However, there is one problem. A rate law for the entire equation must be in terms of its reactants, hydrogen gas and nitric oxide. Therefore, we must reexpress the intermediates dinitrogen dioxide and the hydrogen atom in terms of nitric oxide and molecular hydrogen. One way to do this is to write the reverse rate for the steps with the desired initial reactants. In this case, this will be steps one and two. Rate 1(rev) = k-1N2O2 Rate 1(fwd) = k1NO2 Because step 1 reaches equilibrium, rate 1(rev) = rate 1(fwd) k1NO2 = k-1N2O2 Solve for dinitrogen dioxide k1NO2/k-1 = N2O2 Do the same with step 2 Rate 2(rev) = k-2H4 Rate 2(fwd) = k2H22 Rate (fwd) = Rate(rev) H4 = k2/ k-2 H22 H = (k2/ k-2)1/2H21/2 Plug in these new equations into the overall rate law with intermediates. Rate = kN2O2H Rate = (k2/ k-2)1/2H21/2 * k1NO2/k-1 Rate = kH21/2NO2